All neighbor sum is 0 on a chessboard
$begingroup$
Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.
Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.
What is the fewest 0-valued cells possible after forming the matrix?
mathematics
edited Jan 9 at 20:56
Sesquipedalian
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
$endgroup$
add a comment |
$begingroup$
Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.
Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.
What is the fewest 0-valued cells possible after forming the matrix?
mathematics
edited Jan 9 at 20:56
Sesquipedalian
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
$endgroup$
add a comment |
$begingroup$
Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.
Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.
What is the fewest 0-valued cells possible after forming the matrix?
mathematics
edited Jan 9 at 20:56
Sesquipedalian
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
$endgroup$
Neighbor cells of a matrix are defined as all the cells that share a common edge. A matrix is formed by putting real numbers in each cell.
Form an $8 times 8$ matrix (like a chessboard) where sum of all neighbors of each cell is $0$.
What is the fewest 0-valued cells possible after forming the matrix?
mathematics
mathematics
edited Jan 9 at 20:56
Sesquipedalian
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
edited Jan 9 at 20:56
Sesquipedalian
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
edited Jan 9 at 20:56
Sesquipedalian
357
edited Jan 9 at 20:56
Sesquipedalian
357
edited Jan 9 at 20:56
Sesquipedalian
357
357
asked Jan 7 at 7:46
OrayOray
16.2k437157
asked Jan 7 at 7:46
OrayOray
16.2k437157
asked Jan 7 at 7:46
OrayOray
16.2k437157
16.2k437157
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This works for any given $a,b,c,d$:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$
so:
zero zeroes are needed (for example, $a,b,c,dgt0$)
Following @Jaap's comment:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
add a comment |
$begingroup$
Least amount of zero valued cells is
zero
One such possible grid is:
Because the grid size is even in both directions, I used the following algorithms to fill it:
1. Fill 1 in first two cells. Some other number will also work.
2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).
3. Fill rest of the number in a way that constraints are met.
4. If any of the cells is zero, try with different numbers.
I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This works for any given $a,b,c,d$:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$
so:
zero zeroes are needed (for example, $a,b,c,dgt0$)
Following @Jaap's comment:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
add a comment |
$begingroup$
This works for any given $a,b,c,d$:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$
so:
zero zeroes are needed (for example, $a,b,c,dgt0$)
Following @Jaap's comment:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
add a comment |
$begingroup$
This works for any given $a,b,c,d$:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$
so:
zero zeroes are needed (for example, $a,b,c,dgt0$)
Following @Jaap's comment:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
$endgroup$
This works for any given $a,b,c,d$:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&b&c&d&d&c&b&a\hline -b&-a-c&-b-d&-c-d&-c-d&-b-d&-a-c&-b\hline c&b+d&a+c+d&b+c+d&b+c+d&a+c+d&b+d&c\hline -d&-c-d&-b-c-d&-a-b-c-d&-a-b-c-d&-b-c-d&-c-d&-d\hline d&c+d&b+c+d&a+b+c+d&a+b+c+d&b+c+d&c+d&d\hline -c&-b-d&-a-c-d&-b-c-d&-b-c-d&-a-c-d&-b-d&-c\hline b&a+c&b+d&c+d&c+d&b+d&a+c&b\hline -a&-b&-c&-d&-d&-c&-b&-a\hline end{array}$$
so:
zero zeroes are needed (for example, $a,b,c,dgt0$)
Following @Jaap's comment:
$$begin{array} {|c|c|c|c|c|c|c|c|}hline a&h&c&g&d&f&b&e\hline -h&-a-c&-h-g&-c-d&-f-g&-b-d&-e-f&-b\hline c&g+h&a+c+d&f+g+h&b+c+d&e+f+h&b+d&f\hline -g&-c-d&-f-g-h&-a-b-c-d&-e-f-g-h&-b-c-d&-f-h&-d\hline d&f+g&b+c+d&e+f+g+h&a+b+c+d&f+g+h&c+d&g\hline -f&-b-d&-e-f-g&-b-c-d&-f-g-h&-a-c-d&-g-h&-c\hline b&e+f&b+d&f+h&c+d&g+h&a+c&h\hline -e&-b&-f&-d&-g&-c&-h&-a\hline end{array}$$
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
edited Jan 7 at 13:09
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
answered Jan 7 at 11:03
JonMark PerryJonMark Perry
20.7k64099
20.7k64099
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
add a comment |
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
6
6
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
$begingroup$
You could consider the white squares of the chessboard separately from the black squares, treating them as independent, identical puzzles. You have given those two sets the same (mirrored) solution, but you can choose different $a,b,c,d$ for them. This gives you the most generic parametric solution, with 8 variables on the first row of the board.
$endgroup$
– Jaap Scherphuis
Jan 7 at 12:55
add a comment |
$begingroup$
Least amount of zero valued cells is
zero
One such possible grid is:
Because the grid size is even in both directions, I used the following algorithms to fill it:
1. Fill 1 in first two cells. Some other number will also work.
2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).
3. Fill rest of the number in a way that constraints are met.
4. If any of the cells is zero, try with different numbers.
I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
$endgroup$
add a comment |
$begingroup$
Least amount of zero valued cells is
zero
One such possible grid is:
Because the grid size is even in both directions, I used the following algorithms to fill it:
1. Fill 1 in first two cells. Some other number will also work.
2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).
3. Fill rest of the number in a way that constraints are met.
4. If any of the cells is zero, try with different numbers.
I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
$endgroup$
add a comment |
$begingroup$
Least amount of zero valued cells is
zero
One such possible grid is:
Because the grid size is even in both directions, I used the following algorithms to fill it:
1. Fill 1 in first two cells. Some other number will also work.
2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).
3. Fill rest of the number in a way that constraints are met.
4. If any of the cells is zero, try with different numbers.
I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
$endgroup$
Least amount of zero valued cells is
zero
One such possible grid is:
Because the grid size is even in both directions, I used the following algorithms to fill it:
1. Fill 1 in first two cells. Some other number will also work.
2. Copy the same number to the second end of row and additive inverse in opposite col (or vice versa).
3. Fill rest of the number in a way that constraints are met.
4. If any of the cells is zero, try with different numbers.
I got the above algorithm as an intuition. I am still figuring an easy way to explain/prove why does it work?
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
edited Jan 14 at 6:41
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
answered Jan 7 at 9:23
Mohit JainMohit Jain
2,8701842
2,8701842
add a comment |
add a comment |
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